Question

An electric heater of resistance 8Ω draws 15A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.​

Answer 1

ɢɪᴠᴇɴ:-

Resistane (R) = 8 Ω

Electric current (I) = 15 A

Time (T) = 2 h = 2 × 60 × 60 = 7200 sec

ᴛᴏ ꜰɪɴᴅ:-

Electric Power (P)

ꜱᴏʟᴜᴛɪᴏɴ:-

As we know that,

$\red{\boxed{\gray{\sf{H = I^2 R T}}}}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H\: = (15)^2 \times 8 \times 7200} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H \: = 225 \times 57,600} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H = 12,960,000} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak\green{H \: = \: 12,960,000 \: J} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\tt \pink{\boxed{\purple{\sf{Power = \frac{Heat}{Time}}}}}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {Power = \frac{12960000}{7200}} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {Power = 1800} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak\pink{Power \: = \: 1800 \: Js^{-1}} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

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Answer 2

Given parameters

The resistance of the electric heater (R) =

8 Ω

Current drawn (I) = 15 A

Time = 2 hrs

T = 120 min x 60

Time (T) = 7200 secs

The rate of heat produced by the device in terms of power is expressed as

P= I2R

P = 15 x 15 x 8

P = 1800 watt

Answer

P= 1800 watt

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